Can we use the first equation of motion for this question as they have used second equation of motion to solve it and when I use the first equation of motion I am getting a different answer which is near to it. {answer given =-640m/s^2 : answer I am getting=-648m/s^2}

Q. A train moving at a speed of 180 km/h comes to a stop at a constant acceleration in 15 min after covering a distance of 25 km. What is its acceleration?

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Solution

Dear student,
You can use any of the equation of motion. As all the data is available all three equations provides the answer.
But i don't agree to the answer given
$v = u + a t v = 0 180 k m / h r = 50 m / s 15 m i n = 15 \times 60 = 900 s e c a = - \frac{u}{t} = - \frac{50}{900} a = - 0.055 m / s^{2}$

Regards

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A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hrs longer to cover the total distance. Assuming the uniform speed to be `x` km/h form an equation and solve it to evaluate `x`.

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24

km at a uniform speed. Due to heavy rain its speed gets reduced by

10

km/h and as such it takes two hours longer to cover the total distance. Assuming the uniform speed to be '

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Dear student, You can use any of the equation of motion. As all the data is available all three equations provides the answer. But i don't agree to the answer given v=u+atv=0180km/hr=50m/s15 min =15×60=900seca=-ut=-50900a=-0.055m/s2 Regards